Prove that cos(π/12) − sin(π/12) = 1/√2 Prove that: \[ \cos \frac{\pi}{12} – \sin \frac{\pi}{12} = \frac{1}{\sqrt{2}} \] Solution Using the identity: \[ \cos \theta – \sin \theta = \sqrt{2}\cos\left(\theta+ \frac{\pi}{4}\right) \] Taking \[ \theta = \frac{\pi}{12} \] Then, \[ \cos \frac{\pi}{12} – \sin \frac{\pi}{12} = \sqrt{2}\cos\left(\frac{\pi}{12}+\frac{\pi}{4}\right) \] \[ = \sqrt{2}\cos\left(\frac{\pi}{12}+\frac{3\pi}{12}\right) \] \[ = \sqrt{2}\cos\frac{4\pi}{12} […]

Prove that sin(5π/18) − cos(4π/9) = √3 sin(π/9) Prove that: \[ \sin \frac{5\pi}{18} – \cos \frac{4\pi}{9} = \sqrt{3}\sin \frac{\pi}{9} \] Solution Convert the cosine term into sine form using: \[ \cos \theta = \sin\left(\frac{\pi}{2}-\theta\right) \] \[ \cos \frac{4\pi}{9} = \sin\left(\frac{\pi}{2}-\frac{4\pi}{9}\right) = \sin\frac{\pi}{18} \] Therefore, \[ \sin \frac{5\pi}{18} – \cos \frac{4\pi}{9} = \sin \frac{5\pi}{18} – \sin

Prove that cos 20° + cos 100° + cos 140° = 0 Prove that: \[ \cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0 \] Solution Using the identity: \[ \cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} \] Taking \[ A = 20^\circ,\qquad B = 100^\circ \] Then, \[ \cos

Prove that cos 80° + cos 40° − cos 20° = 0 Prove that: \[ \cos 80^\circ + \cos 40^\circ – \cos 20^\circ = 0 \] Solution Using the identity: \[ \cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} \] Taking \[ A = 80^\circ,\qquad B = 40^\circ \] Then, \[ \cos

Prove that sin 50° − sin 70° + sin 10° = 0 Prove that: \[ \sin 50^\circ – \sin 70^\circ + \sin 10^\circ = 0 \] Solution Using the identity: \[ \sin A – \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} \] Taking \[ A = 50^\circ,\qquad B = 70^\circ \] Then, \[ \sin

Prove that cos 55° + cos 65° + cos 175° = 0 Prove that: \[ \cos 55^\circ + \cos 65^\circ + \cos 175^\circ = 0 \] Solution Using the identity: \[ \cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} \] Taking \[ A = 55^\circ,\qquad B = 65^\circ \] Then, \[ \cos

Prove that sin 40° + sin 20° = cos 10° Prove that: \[ \sin 40^\circ + \sin 20^\circ = \cos 10^\circ \] Solution Using the identity: \[ \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \] Taking \[ A = 40^\circ,\qquad B = 20^\circ \] Then, \[ \sin 40^\circ + \sin 20^\circ

Prove that sin 105° + cos 105° = cos 45° Prove that: \[ \sin 105^\circ + \cos 105^\circ = \cos 45^\circ \] Solution Using the identity: \[ \sin \theta + \cos \theta = \sqrt{2}\sin(\theta+45^\circ) \] Taking \[ \theta = 105^\circ \] Then, \[ \sin 105^\circ + \cos 105^\circ = \sqrt{2}\sin(105^\circ+45^\circ) \] \[ = \sqrt{2}\sin 150^\circ

Prove that sin 23° + sin 37° = cos 7° Prove that: \[ \sin 23^\circ + \sin 37^\circ = \cos 7^\circ \] Solution Using the identity: \[ \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \] Taking \[ A = 23^\circ,\qquad B = 37^\circ \] Then, \[ \sin 23^\circ + \sin 37^\circ

Prove that sin 50° + sin 10° = cos 20° Prove that: \[ \sin 50^\circ + \sin 10^\circ = \cos 20^\circ \] Solution Using the identity: \[ \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \] Taking \[ A = 50^\circ,\qquad B = 10^\circ \] Then, \[ \sin 50^\circ + \sin 10^\circ